Nair S. Advanced Topics in Applied Mathematics: For Engineering and the Physical Sciences

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Green’s Functions 29

in a three dimensional (3D) domain  with homogeneous conditions

on the boundary ∂. Here, p and q are functions of x, y,andz. We will

also consider the two-dimensional version where the z-dependence is

absent. Deﬁning the gradient operator

∇ = i

∂

∂x

∂

∂y

∂

∂z

, (1.159)

where i , j ,andk are the cartesian unit vectors, we can write the

Sturm-Liouville equation as

∇ ·(p∇u) +qu =f . (1.160)

Let

n =in

+j n

+kn

(1.161)

be the outward normal to the boundary surface ∂.

The inner product is now deﬁned as the volume integral

u,v =





uvd. (1.162)

As before, let the Green’s function, g(x, ξ), satisfy

Lg = δ(x −ξ). (1.163)

Then

g,Lu−u, Lg=0 =g, f −u,δ, (1.164)

provided g satisﬁes the homogeneous boundary conditions on ∂.We

now have

u(x) =





g(x, ξ)f (ξ )d. (1.165)

Formally, this shows the Green’s function representation of the solu-

tion of any self-adjoint partial differential equation in an n-dimensional

domain. The extension of this representation for any non-self-adjoint

operator is similar to what was done in the case of the one-dimensional

equations.

30 Advanced Topics in Applied Mathematics

To see the behavior of the Green’s function in the neighborhood of

the source point ξ , we consider a small spherical volume, V, of radius

r = centered at ξ and integrate the equation, Lu =δ, over this volume.

Using the fact that the integral of the delta function is unity, we see



[∇ ·(p∇g) +qg]dV =1. (1.166)

Using the Gauss theorem, the ﬁrst term of the integral can be converted

to a surface integral,



n ·p∇ gdS+



qg dV = 1,



∂g

∂r

dS +



qg dV = 1, (1.167)

where we have used the directional derivative

n ·p∇g = p

∂g

∂r

, (1.168)

with r representing a coordinate normal to the surface (i.e., r =|x −ξ|).

As  →0, we observe that the volume integral of qg is small compared

to the surface integral, and the Green’s function is spherically symmet-

rical as the boundaries are far away compared to the scale of . From

this the behavior of the Green’s function for small values of r is of the

form,

4πpr

∼1,

∼

4πpr

. (1.169)

For an inﬁnite domain the boundary effects are absent, and, in addition,

if q =0, the preceding result can be written as

4πpr

=1,

4πpr

. (1.170)

Further, if p = 1, the exact Green’s function for an inﬁnite domain

becomes

∞

=−

4πr

, r ={(x −ξ)

+(y −η)

+(z −ζ)

}

1/2

. (1.171)

Green’s Functions 31

(ξ, η)

Figure 1.8. Two-dimensional domain.

With p = 1andq = 0, the Sturm-Liouville equation becomes the

Poisson equation

∇

u =f . (1.172)

We could apply the above integration using the Gauss theorem for

the two-dimensional (2D) Sturm-Liouville equation (see Fig. 1.8). This

results in

2πpr

∼1,

∼

2πpr

. (1.173)

Again, when p =1, q =0 we have the Poisson equation,

∇

u =f , (1.174)

with the exact Green’s function for the inﬁnite domain,

∞

2π

logr, r ={(x −ξ)

+(y −η)

}

1/2

. (1.175)

Now we have exact Green’s functions for the Laplace operators in 2D

and 3D inﬁnite spaces. To obtain the solution u in terms of g

∞

, we need

to compute the integrals of f multiplied by g over the whole space. For

these integrals to exist, certain conditions on the decay of f at inﬁnity

are required. Of course, in bounded domains, g

∞

does not satisfy the

boundary conditions, and we have to resort to other methods.

32 Advanced Topics in Applied Mathematics

1.11.1 Example: Steady-State Heat Conduction in a Plate

Consider an inﬁnite plate under steady-state temperature distribution

with a heat source distribution, q(x,y). The temperature, T, satisﬁes

∇

T =−

, (1.176)

where k is the conductivity. Using the two-dimensional Green’s

function, the solutions is written as

T(x,y) =−

4πk



∞

−∞



∞

−∞

q(ξ, η)log[(x −ξ)

+(y−η)

]dξ dη. (1.177)

Usually, the source is limited to a ﬁnite area, and the limits of the

above integral will have ﬁnite values. If the heat source has a circular

boundary, polar coordinates may be more convenient.

1.11.2 Example: Poisson’s Equation in a Rectangle

To solve the Poisson’s equation

∇

u =f (x, y); −a < x < a, −b < y < b, (1.178)

with u = 0 on the rectangular boundary, we need the Green’s func-

tion satisfying the same boundary condition. Here, we illustrate the

eigenfunction representation of the Green’s function for this pur-

pose. Let u

(where m and n will take integer values) represent an

eigenfunction of the Laplace equation with u

=0 on the boundary.

We have

∂u

∂x

∂u

∂y

=λ

. (1.179)

Using separation of variable, we represent u

(x,y) =X

(x)Y

(y). (1.180)

Substituting this in the Laplace equation and dividing everything by

,weget



=λ

. (1.181)

Green’s Functions 33

Let



=−µ



=−ν

, λ

=−(µ

+ν

). (1.182)

Solutions of these equations with X

(±a) =0andY

(±b) =0are

=sinmπx/a, Y

=sinnπy/b; µ

=mπ/a, ν

=nπ/b.

(1.183)

By integrating these functions over their respective intervals, we can

make their norms unity if we scale these as

√

sin

mπx

, Y

√

sin

nπy

. (1.184)

The two-dimensional eigenfunctions and eigenvalues are

(x,y) =

√

sin

mπx

sin

nπy

, λ

=−π





. (1.185)

The Green’s function becomes

g(x, y,ξ ,η) =

∞



m=1

∞



n=1

(x,y)u

(ξ, η)

. (1.186)

The solution of the nonhomogeneous equation using this Green’s func-

tion is identical to the one we could obtain using a Fourier series

approach.

1.11.3 Steady-State Waves and the Helmholtz Equation

The wave equation for a ﬁeld variable, , propagating with a speed c

is given by

∇

 =

∂



∂t

. (1.187)

If the waves are harmonic, we can separate the time dependence and

space dependence in the form

 =ψ(x,y, z)e

−it

, (1.188)

34 Advanced Topics in Applied Mathematics

where  is the angular frequency, and ψ satisﬁes

∇

ψ +k

ψ = 0, k = /c, (1.189)

with k representing the wave number. This equation is called the

Helmholtz equation. From our discussion of the Sturm-Liouville prob-

lem, now we have p = 1andq = k

, and the singular behavior of the

Green’s function is unaffected. By direct substitution, it can be veriﬁed

that

g =−

ikr

4πr

(1.190)

is the required Green’s function for the outgoing waves (the exponent,

kr −t,in can be kept constant if r and t both increase by r and

r/c, respectively).

For the 2D case,

g =−

(1)

(kr), (1.191)

which is one of the Hankel functions

(1)

+iY

, H

(2)

−iY

, (1.192)

with J

and Y

being the Bessel and Neumann functions, respectively.

The Hankel functions have logarithmic behavior through the Neu-

mann function Y

,asr →0. We use H

(1)

for outgoing waves and H

(2)

for incoming waves.

1.12 METHOD OF IMAGES

We could use the Green’s function, g

∞

, for the 2D inﬁnite domain,

obtained here to solve the Laplace equation in other domains, in two

ways: one is the method of images, which is useful if the new domain

can be obtained by symmetrically folding the full inﬁnite domain and

the other is through conformal mapping.

As shown in Fig. 1.9, to construct the Green’s function for the semi-

inﬁnite domain, x > 0, with g =0atx =0, we introduce the usual source

Green’s Functions 35

(ξ,η)

(−ξ,η)

Figure 1.9. Semi-inﬁnite domain with a source and its image.

at (ξ, η) and an image sink at (−ξ ,η). The sink is outside the domain,

but the combined effect is to have

g =

4π

Log

(x −ξ)

+(y −η)

(x +ξ)

+(y −η)

, (1.193)

equals to zero on x = 0. Here, we have extended g

∞

into x < 0inan

odd fashion. If we needed g with normal derivative zero on x = 0, we

have to extend g

∞

in an even fashion, using two sources.

To obtain the Green’s function for a quarter plane we need four

sources (two of them may be sinks depending on the boundary condi-

tions). Similarly, for a 45

◦

wedge, we use eight sources at the points,

(ξ, η), (η, ξ)and at the images of these two points under reﬂection with

respect to the x-andy-axes.

We can even obtain the Green’s function for a rectangle with one

corner at (0, 0) and the diagonally opposite corner at (a,b) by repeat-

edly reﬂecting the original source at (ξ ,η) about (a) the x = 0 line, (b)

x = a line, (c) y = 0 line, and (d) y = b line. This gives a doubly inﬁnite

system of sources.

For the 3D case, if the new domain can be obtained by symmetri-

cally sectioning the full inﬁnite domain, we can construct the Green’s

function by the method of images.

36 Advanced Topics in Applied Mathematics

1.13 COMPLEX VARIABLES AND THE LAPLACE EQUATION

The real and imaginary parts of an analytic function of a complex

variable automatically satisfy the Laplace equation in 2D. Also, using

the conformal mapping, it is possible to map different domains into

domains with convenient boundary curves. First, let us note that

2πg

∞

=Re[Log(z)]=Log |z|=Log r, (1.194)

satisﬁes

∇

∞

=δ(x, y). (1.195)

We can move the source to ζ = ξ +iη by deﬁning

2πg

∞

=Log |z −ζ |. (1.196)

For a semi-inﬁnite domain, 0 < y, with g = 0 on the real axis y =0,

using the method of images, we introduce a sink at ξ −iη =

ζ and write

2πg =Log



z −ζ

z −



. (1.197)

We can map (see Fig. 1.10) the upper half plane onto the interior of

a unit circle, with the line y =0 becoming the circular boundary, using

w =

i −z

i +z

,orz =

w −1

w +1

. (1.198)

1/ ¯

Figure 1.10. Mapping the upper half plane to the interior of a circle.

Green’s Functions 37

Let α be the image of the source point ζ . Then

ζ =

α −1

α +1

. (1.199)

Using these in the expression for g, after some simpliﬁcation, we get

2πg =Log



w −α

1 −¯αw



. (1.200)

Now we have the Green’s function for a unit circle with g being zero

on the boundary.

Another way to view this Green’s function is to consider

2πg =Log |z|, (1.201)

as the solution for the Laplace equation with the right type of

singularity at z =0. The transformation

z =e

iφ

w −α

1 −¯αw

(1.202)

moves the origin to α but retains the circular boundary. The angle φ

gives a rigidbody rotation.

Using the notation

w = re

iθ

, α = ρe

iφ

, (1.203)

we can write

g(r, θ,ρ,φ) =

4π

Log

−2rρ cos(φ −θ)+ρ

1 −2rρ cos(φ −θ)+r

, (1.204)

which solves the Laplace equation on a unit circle. In this form, it is

easy to see that g is indeed zero when r = 1.

Conformal mapping can be used to map domains onto a unit cir-

cle and the Green’s function, Eq. (1.204), can be used to solve the

Poisson equation. In particular, the Schwartz-Christoffel transform

maps polygons onto the upper half plane.

38 Advanced Topics in Applied Mathematics

1.13.1 Nonhomogeneous Boundary Conditions

Consider the Poisson equation

∇

u =f , (x,y) ∈, (1.205)

with the boundary condition

u =h, (x, y) ∈ ∂. (1.206)

Let g satisfy

∇

g = δ(x −ξ,y −η), g = 0on(x,y) ∈∂. (1.207)

The inner products give

g,∇

u−u,∇

g=





∂u

∂n

−u

∂g

∂n



ds. (1.208)

As g = 0 on the boundary, the ﬁrst term on the right is zero, and we

ﬁnd

u(ξ , η) =





g(x, y,ξ ,η)f (x,y) dxdy +



∂g

∂n

ds. (1.209)

As long as g =0 on the boundary, we can incorporate nonhomogeneous

boundary conditions without any complications.

1.13.2 Example: Laplace Equation in a Semi-inﬁnite Region

To solve

∇

u =0, −∞ < x < ∞,0< y < ∞, (1.210)

with u =f (x) on the boundary y =0, we use

g =

2π

Log



z −ζ

z −



. (1.211)

Assuming u tends to zero at inﬁnity, Eq. (1.209) becomes

u(ξ , η) =

2πi



∞

−∞



z −ζ

−

z −



f (x)dx, y =0. (1.212)